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Math Help - Ordinary differential equations

  1. #1
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    Ordinary differential equations

    Hi there. Well, I have some doubts about this exercise, I think I've solved it, but I wanted your opinion, which always help. So, it says:

    There is no general method for solving the homogeneous general equation of second order
    y''+P(x)y'+Q(x)y=0 (1)

    But if we already know a solution y_1(x)\neq 0, then we always can find a second solution linearly independent

    y_2(x)=y_1(x) \int \dysplaystyle\frac{e^{-\int P(x)dx}}{y_1^2}dx
    Demonstrate that this functions form a basis for the space of solutions of the differential equation (1).


    So, what I did is simple. I know that if the solution is linearly independent, then the Wronskian w(y_1,y_2,x) must be zero. So, I just made the calculus for the wronskian:

    \left| \begin{matrix}{y_1}&{y_2}\\{y_1'}&{y_2'}\end{matri  x} \right|=\left| \begin{matrix}{y_1}&{y_1(x) \int \dysplaystyle\frac{e^{-\int P(x)dx}}{y_1^2}dx}\\{y_1'}&{y_1'(x) \int \displaystyle\frac{e^{- \int P(x)dx}}{y_1^2}dx+y_1(x) \displaystyle\frac{e^{- \int P(x)dx}}{y_1^2(x)}dx}\end{matrix} \right|=\displaystyle\frac{e^{- \int P(x)dx}}{y_1(x)}\neq0

    Is this right?

    Bye there.
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  2. #2
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    Your approach is fine; however I don't think your determinant computation is correct. You've set it up properly, but I get just

    e^{-\int P(x)\,dx}

    for the determinant.
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  3. #3
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    You're right! thanks.
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  4. #4
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    You're welcome!
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  5. #5
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    This is pretty much like the same. It gives me a differential equation, and one solution and asks me for the entire solution. Did I do it well?

    (1-2x)y''+2y'+(2x-3)y=0,y_1=e^x
    y_2=e^x \int \displaystyle\frac{e^{-\int 2dx}}{e^{2x}}dx=\displaystyle\frac{e^{-3x}}{-4}

    Then y(x)=Ae^x-B\displaystyle\frac{e^{-3x}}{4}

    Is that okey? I'm not sure, because I've tried to corroborate the solution with wolfram alpha, but it gives something different.

    Here:
    (1-2x)y''+2y'+(2x-3)y=0 - Wolfram|Alpha
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  6. #6
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    You haven't normalized the DE to look like (1) in the OP. You have to do that before applying the variation of parameters formula.
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  7. #7
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    Sorry, I've missed that part, how should I do that?
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  8. #8
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    Just divide the entire DE by 1-2x to get the coefficient of y'' equal to 1.
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